# Quadratic Formula



## Lorem Ipsum (Oct 9, 2010)

I've been asked by my maths teacher to teach the class the quadratic formula on Monday. And to make things more informative I've decided to show how to derive the original equation. Now, these are the workings I've got, but I just want to know whether they're correct or not so I don't make a fool out of myself.

Here goes:

ax²+bx+c = 0

x² + (b/a)x + c/a = 0/a = 0 - Dividing by the coefficient of the square term

(x+b/2a)(x+b/2a) + c/a = 0 - Dividing the "b" coefficient by two and factorising

[x² + 2bx/2a + b²/4a²] - Expanding the brackets to see the difference between this and the original equation

[x² + bx/a + b²/4a²] - Simplifying

(x + b/2a)² - b²/4a² + c/a = 0 - Completing the square

x+ b/2a = (±√b²-4ac)/2a - Rearrangement

x = (-b±√b²-4ac)/2a - Final equation


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## Butterfree (Oct 9, 2010)

> (x+b/2a)(x+b/2a) + c/a = 0 - Dividing the "b" coefficient by two and factorising


Huh? You can't just decide to divide one of the coefficients in an equation by 2. o_O The supposed factorization doesn't even multiply to x^2 + (b/a)x. I don't follow.


I don't remember how it was actually taught to me, but reverse engineering gets me this:

ax² + bx + c = 0

Multiply both sides by 4a: 4a²x² + 4abx + 4ac = 0

Rearrange: 4a²x² + 4abx = -4ac

Complete the square: 4a²x² + 4abx + b² = b² - 4ac

Factorize: (2ax + b)² = b² - 4ac

Take the square root of both sides: 2ax + b = ±√(b² - 4ac)

Subtract b on both sides: 2ax = -b ±√(b² - 4ac)

Divide by 2a: x = (-b ±√(b² - 4ac))/2a


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## Lorem Ipsum (Oct 9, 2010)

When completing the square, normally you divide the "b" coefficient by two so when you factorise it, it makes sense. That was such a poor explanation it's untrue: it's better to explain:

x² + 8x + 13 = 0
_This can't be perfectly factorised with whole digits, so complete the square_

(x+4)(x+4)
= x² + 8x + 16 = 3
_There's a difference of three between this and the original equation._

(x+4)² = 3
_So we simplify the formula and take the three to make it balanced again_

x = -4±√3
_This gives the two values of x because it's either plus or minus_

The reason you divide by two is so you have an easy value to put in the factorised brackets [(x+4)(x+4) in this case] and you just take the difference then balance the equation.


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## Music Dragon (Oct 9, 2010)

Show, don't tell!

... Hurr. No, but in all seriousness; if you want to be truly informative, then I think it would help to actually show how the "square" is completed with an illustration or something - either that, or take it very, very slowly. (Or both!)

To most people, a wall of numbers says about as much as a wall of text.


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## Butterfree (Oct 9, 2010)

That makes sense, but (x+b/2a)(x+b/2a) + c/a = 0 as an equation is simply incorrect; if you're going to complete the square, you have to actually add the term on _both_ sides instead of just writing up an incorrect equation and then subtracting the term you added a few steps later. _First_ add b²/4a² on both sides, _then_ factorize. And preferably, write your logic for choosing to add b²/4a² somewhere off at the side, where you aren't equating it with anything.


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## Aobaru (Oct 9, 2010)

Bad memories from Algebra 2. Anyway, you're not completing the square correctly.

You have to subtract _c_ to the other side of the equation, first.  *ax² + bx = -c*

Divide by coefficient of _x²_:  *x² + bx/a = -c/a*

Add the square of half the coefficient of _x_ to both sides:  *x² + bx/a + (b/2a)² = (b/2a)² - c/a*

Now the left side can be written as a binomial, and the right side can be simplified.  *(x + b/2a)² = (b² - 4ac)/4a²*

Take the square root of both sides.  *x + b/2a = √[(b² - 4ac)/4a²]*

Finally, subtract _b/2a_ from both sides, and you get: * x = [-b±√(b²-4ac)]/2a *


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## opaltiger (Oct 10, 2010)

Here's my method, for the record:

ax² + bx + c = 0
a(x² + (b/a)x) + c = 0
a((x + (b/2a))² - (b²/4a²)) + c = 0
a(x + (b/2a))² - (ab²/4a²) + c = 0
a(x + (b/2a))² = (b²/4a) - c
4a²(x + (b/2a))² = b² - 4ac
(x + (b/2a))² = (b² - 4ac)/4a²
x + (b/2a) = ±√(b² - 4ac)/2a
x = (-b ± √(b² - 4ac))/2a


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## Murkrow (Oct 15, 2010)

Sorry for this not being entirely relevant to the thread but it involves the quadratic formula and stuff so I'll post it here, instead of bumping that other maths thread from last year. Also sorry for probably being awful at explaining things well over text. This was originally going to be a question asking for help but I worked it out in the middle of writing the post, heh. (but then more questions came up but then I worked them out but then...)

Someone from the year below me asked for some help with his maths homework. His question was:
The sum of two of a rectangle's sides is S, the area is A. Prove that S² ≥ 4A. (Suggestion: The rectangle's sides are x and S-x)

Since the homework itself involved the quadratic formula, I assume there's some way to prove it using that. However the only way I could get an answer at the time was by doing this

x(S-x) = A
-x² + Sx = A
x² - Sx = -A
(x - S/2)² - S²/4 = -A
4(x - S/2)² - S² = -4A

His class haven't done complex numbers so (x - S/2)² ≥ 0, and so S² which is being taken away from this has to be greater than or equal to 4A for it to make -4A.

I'm sure there's some other way to work it out though, since I remember doing the same homework last year. I then realised the rest of the questions on that homework sheet were proving that a quadratic equation did/n't have real roots by working out b² - 4ac

This (much simpler!) way obviously works:
x² - Sx + A = 0

b² - 4ac ≥ 0
S² - 4A ≥ 0 (I realise that it's (-S)² but I usually just skip that step if it's negative)
S² ≥ 4A

This was the reason I was going to make this post in the first place since I didn't see why it needed real roots as I had no idea what the roots were in the context of the rectangle.

EDIT: Removed this part since I did a mistake, hold on a second!

Now I realise that if you were to arrange four rectangles like this:






Then the green area is 4A, and S² is the entire area of the square. So S² - 4A is the area of the white square in the middle - and since you can't have negative area, it must be 0 or more. This makes sense and answers my question.

However, I also realise that the difference between the width and height of the rectangles is the same as the sides of the white square. Which is x (and it would make sense that x is real, so that's another, probably more obvious, answer to my previous question).  So:

x = ± √(S² - 4A)
But using the quadratic formula,
x² - Sx + A = 0
x = (S ± √(S² - 4A))/2

Which is a different answer. Where did I go wrong?

EDIT2: Noticed where I went wrong, I originally labelled the sides of the rectangles in the picture as S - x and S. Instead of S - x and x. So the side of the white square is -S + 2x and not x.

-S + 2x = ± √(S² - 4A)
2x = S ± √(S² - 4A)
x = (S ± √(S² - 4A))/2

Which is the answer I was looking for.


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